Contributed by:
The highlights are:
1. Internal Energy
2. Work
3. Heating
4. Energy Conservation – the First Law
5. Quasi-static processes
6. Enthalpy
7. Heat Capacity
1.
Lecture 2 The First Law of
Thermodynamics (Ch.1)
1. Internal Energy, Work, Heating
2. Energy Conservation – the First Law
3. Quasi-static processes
4. Enthalpy
5. Heat Capacity
2.
Internal Energy
The internal energy of a system of particles, U, is the sum of the kinetic
energy in the reference frame in which the center of mass is at rest and the
potential energy arising from the forces of the particles on each other.
system Difference between the total energy and the internal energy?
boundary system
U = kinetic + potential
B The internal energy is a state function – it depends only on
P
the values of macroparameters (the state of a system), not
on the method of preparation of this state (the “path” in the
T V macroparameter space is irrelevant).
A
In equilibrium [ f (P,V,T)=0 ] : U = U (V, T)
U depends on the kinetic energy of particles in a system and an average
inter-particle distance (~ V-1/3) – interactions.
For an ideal gas (no interactions) : U = U (T) - “pure” kinetic
3.
Internal Energy of an Ideal Gas
The internal energy of an ideal gas f
with f degrees of freedom:
U Nk BT
2
f 3 (monatomic), 5 (diatomic), 6 (polyatomic)
(here we consider only trans.+rotat. degrees of freedom, and neglect
the vibrational ones that can be excited at very high temperatures)
How does the internal energy of air in this (not-air-tight) room change
with T if the external P = const?
f PV f
U N in room k BT N in room PV
2 k BT 2
- does not change at all, an increase of the kinetic energy of individual
molecules with T is compensated by a decrease of their number.
4.
Work and Heating (“Heat”)
WORK
We are often interested in U , not U. U is due to:
Q - energy flow between a system and its
environment due to T across a boundary and a finite HEATING
thermal conductivity of the boundary
– heating (Q > 0) /cooling (Q < 0)
(there is no such physical quantity as “heat”; to
emphasize this fact, it is better to use the term
“heating” rather than “heat”)
W - any other kind of energy transfer across
boundary - work
Work and Heating are both defined to describe energy transfer
across a system boundary.
Heating/cooling processes:
conduction: the energy transfer by molecular contact – fast-moving
molecules transfer energy to slow-moving molecules by collisions;
convection: by macroscopic motion of gas or liquid
radiation: by emission/absorption of electromagnetic radiation.
5.
The First Law
The first law of thermodynamics: the internal energy of a system can be
changed by doing work on it or by heating/cooling it.
U = Q + W conservation of energy.
Sign convention: we consider Q and W to be positive if energy
flows into the system.
P
For a cyclic process (Ui = Uf) Q = - W.
If, in addition, Q = 0 then W = 0 V
T
An equivalent formulation:
Perpetual motion machines of the first type do not exist.
6.
Quasi-Static Processes
Quasi-static (quasi-equilibrium) processes – sufficiently
slow processes, any intermediate state can be considered
as an equilibrium state (the macroparamers are well-
defined for all intermediate states).
Advantage: the state of a system that participates in a quasi-equilibrium
process can be described with the same (small) number of macro
parameters as for a system in equilibrium (e.g., for an ideal gas in quasi-
equilibrium processes, this could be T and P). By contrast, for non-
equilibrium processes (e.g. turbulent flow of gas), we need a huge number
of macro parameters.
Examples of quasi- For quasi-equilibrium processes, P, V, T are
equilibrium processes: well-defined – the “path” between two states is
a continuous lines in the P, V, T space.
isochoric: V = const P 2
isobaric: P = const
isothermal: T = const
1 V
adiabatic: Q=0 T
7.
Work
The work done by an external force on a gas
A – the enclosed within a cylinder fitted with a piston:
piston
area W = (PA) dx = P (Adx) = - PdV
force
The sign: if the volume is decreased, W is positive (by
x
compressing gas, we increase its internal energy); if the
volume is increased, W is negative (the gas decreases
its internal energy by doing some work on the
P environment).
V2
W1 2 P (T , V )dV
V1
W = - PdV - applies to any
shape of system boundary dU = Q – PdV
The work is not necessarily associated with the volume changes – e.g.,
in the Joule’s experiments on determining the “mechanical equivalent of
heat”, the system (water) was heated by stirring.
8.
W and Q are not State Functions
V2 - we can bring the system from state 1 to
W1 2 P (T , V )dV state 2 along infinite # of paths, and for each
V1
path P(T,V) will be different.
P 2
Since the work done on a system depends not
only on the initial and final states, but also on the
1 V intermediate states, it is not a state function.
T
U = Q + W U is a state function, W - is not
thus, Q is not a state function either.
P
A B Wnet WAB WCD P2 V2 V1 P1 V1 V2
P2 P1 V2 V1 0
P1 D C
- the work is negative for the “clockwise” cycle; if
the cyclic process were carried out in the reverse
V1 V2 V order (counterclockwise), the net work done on
the gas would be positive.
PV diagram
9.
Comment on State Functions
U, P, T, and V are the state functions, Q and W are not. Specifying an initial and final
states of a system does not fix the values of Q and W, we need to know the whole
process (the intermediate states). Analogy: in classical mechanics, if a force is not
conservative (e.g., friction), the initial and final positions do not determine the work, the
entire path must be specified.
In math terms, Q and W are not exact differentials of some functions
of macroparameters. To emphasize that W and Q are NOT the state U
functions, we will use sometimes the curled symbols (instead of d)
for their increments (Q and W).
V
d U T d S P dV - an exact differential S
y z(x1,y1) dz Ax x, y dx Ay x, y dy - it is an exact differential if it is
the difference between the values of some (state) function
z(x2,y2) dz z x dx, y dy z x, y
z(x,y) at these points:
x Ax x, y Ay x, y
A necessary and sufficient condition for this:
y x
If this condition z x, y z x, y z z
Ax x, y Ay x, y d z dx dy
holds: x y x y y x
f T
e.g., for an ideal gas: Q dU PdV Nk B dT dV - cross derivatives
2 V are not equal
10.
Problem
Imagine that an ideal monatomic gas is taken from its initial state A to state
B by an isothermal process, from B to C by an isobaric process, and from
C back to its initial state A by an isochoric process. Fill in the signs of Q,
W, and U for each step.
P,
105 Pa Step Q W U
A
2
AB + -- 0
T=const
BC -- + --
B
1 C
CA + 0 +
1 2 V, m3
f PV Nk B T
U Nk BT
2
11.
Quasistatic Processes in an Ideal Gas
isochoric ( V = const )
2 W1 2 0
PV= NkBT2
3
Q1 2 Nk B T2 T1 0 CV T
1 PV= NkBT1 2
(see the last slide)
V1,2 V dU Q1 2
isobaric ( P = const )
P 2
W1 2 P(V , T )dV PV2 V1 0
2 1
1 PV= NkBT2 5
PV= NkBT1 Q1 2 Nk B T2 T1 0 CP T
2
V1 V2 V
dU W1 2 Q1 2
12.
Isothermal Process in an Ideal Gas
P isothermal ( T = const ) :
PV= NkBT
dU 0
W V2 V2
dV V
V2 V1 W1 2 P (V , T )dV Nk BT Nk BT ln 2
V V V1
V1 V1
V Q1 2 W1 2
Wi f Nk BT ln i
Vf
Wi-f > 0 if Vi >Vf (compression)
Wi-f < 0 if Vi
13.
Adiabatic Process in an Ideal Gas
adiabatic (thermally isolated system) Q1 2 0 dU W1 2
The amount of work needed to change the state of a thermally isolated system
depends only on the initial and final states and not on the intermediate states.
V2
W1 2 P(V , T )dV
P V1
2
to calculate W1-2 , we need to know P (V,T)
for an adiabatic process
PV= NkBT2
1 f f
PV= NkBT1 U Nk BT dU Nk B dT PdV
2 2
V2 V1 V ( f – the # of “unfrozen” degrees of freedom )
2
PV Nk BT PdV VdP Nk B dT PdV VdP PdV PV
f
V P
dV 2 dP 2 Adiabatic dV dP
1 0 , 1 0
V f P f exponent V1
V P1 P
V P
ln ln 1 PV P1V1 const
V1 P
14.
Adiabatic Process in an Ideal Gas (cont.)
P PV P1V1 const
2
An adiabata is “steeper” than an isotherma:
PV= NkBT2 in an adiabatic process, the work flowing
1 out of the gas comes at the expense of its
PV= NkBT1
thermal energy its temperature will
V2 V1 V decrease.
V2 V2 V2
PV 1 1
1 1
W1 2 P(V , T )dV dV PV
1 1
V
V1 V1
V
1 V1
1 1 1
PV1 1
1 1
1 V2 V1
1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 1.33 (polyatomic)
(again, neglecting the vibrational degrees of freedom)
f f
Prove W1 2 PV
Nk B T U
2 2
15.
Summary of quasi-static processes of ideal gas
U U f U i
Quasi-Static Ideal gas
U Q W
process law
isobaric f f f 2 Vi V f
U Nk B T P V P V P V
(P=0) 2 2 2 Ti T f
isochoric f f f Pi Pf
U Nk B T P V P V 0
(V=0) 2 2 2 Ti T f
isothermal Vf
0 W Nk B T ln PV
i i Pf V f
(T=0) Vi
adiabatic f f
U Nk B T PV PV
i i Pf V f
0 U
(Q=0) 2 2
16.
Problem
Imagine that we rapidly compress a sample of air whose initial pressure is
105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of
its original volume (e.g., pumping bike’s tire). What is its final temperature?
Rapid compression – approx. adiabatic, no time for the energy
exchange with the environment due to thermal conductivity
P1V1 Nk BT1 1
PV
PV
PV V1 T2
P2V2 Nk BT2 P2 1 1 1 1
Nk BT2 1 1 T2
1
V2
V2 T1 V2 T1
P1V1 P2V2
1 1
For adiabatic processes: T1 V 1 T2 V2 const
also P 1 / T const
1
V1
T2 T1 295 K 40.4 295 K 1.74 514 K
V2
- poor approx. for a bike pump, works better for diesel engines
17.
Non-equilibrium Adiabatic Processes
Free expansion
1. TV 1 const V – increases
T – decreases (cooling)
2. On the other hand, U = Q + W = 0
U ~ T T – unchanged
(agrees with experimental finding)
Contradiction – because approach
#1 cannot be justified – violent
expansion of gas is not a quasi-
static process. T must remain the
same.
TV 1 const - applies only to quasi-equilibrium processes !!!
18.
The Enthalpy
Isobaric processes (P = const):
dU = Q - PV = Q -(PV) Q = U + (PV)
H U + PV - the enthalpy
The enthalpy is a state function, because U, P,
and V are state functions. In isobaric processes,
the energy received by a system by heating equals
to the change in enthalpy.
in both cases, Q
isochoric: Q=U does not
depend on the
isobaric: Q=H path from 1 to 2.
Consequence: the energy released (absorbed) in chemical reactions at constant
volume (pressure) depends only on the initial and final states of a system.
f f
The enthalpy of an ideal gas: H U PV Nk BT Nk BT 1 Nk BT
(depends on T only) 2 2
19.
Heat Capacity
The heat capacity of a system - the amount of energy
Q
transfer due to heating required to produce a unit C
temperature rise in that system T
T
f1 f2 f3
C is NOT a state function (since Q is not a T1+dT
state function) – it depends on the path
between two states of a system T1
i
V
( isothermic – C = , adiabatic – C = 0 )
C
The specific heat capacity c
m
20.
CV and CP
nst U the heat capacity at
= c o CV
Q dU PdV
V T V constant volume
C P=
dT dT con
st
H the heat capacity at
C P
T P constant pressure
To find CP and CV, we need f (P,V,T) = 0 and U = U (V,T)
f f
For an ideal gasU Nk BT H 1 Nk BT
2 2
f f f
CV Nk B nR CP 1 nR
2 2 2
# of moles
For one mole of a 3 5
CV R C P R
monatomic ideal gas: 2 2
21.
Another Problem P
Pi
During the ascent of a meteorological helium-gas filled balloon,
its volume increases from Vi = 1 m3 to Vf = 1.8 m3, and the
pressure inside the balloon decreases from 1 bar (=10 5 N/m2) to Pf
0.5 bar. Assume that the pressure changes linearly with volume
between Vi and Vf.
(a) If the initial T is 300K, what is the final T?
(b) How much work is done by the gas in the balloon? Vi Vf V
(c) How much “heat” does the gas absorb, if any?
PV 0.625 bar/m3 V 1.625 bar
(a) PV Nk T PV Pf V f 0.5bar 1.8m3
B T T f Ti 300K 3
270K
Nk B Pi Vi 1bar 1m
Vf Vf
(b) WON P (V )dV - work done on a system WBY P (V )dV - work done by a system
Vi Vi
Vf
WON WBY WBY P(V )dV 0.5 0.8 bar m3 0.5 0.4 bar m3 0.6bar m3 6 10 4 J
Vi
(c) U Q WON
3 3 T
Q U WON NkB T f Ti WON Pi Vi f 1 WBY 1.5 105 J 0.1 6 104 J 4.5 104 J
2 2 Ti
+1-315-636-4485
+91 98151-74050 
